Python: Would it be possible to call a function specifying by string variable?

Question

Given there is a list of function's names as strings, would it be possible to call corresponding function with random sampling from the list? Currently I am hard coding all the name by if statement so that

def a(x):
    print(x)

def b(y):
    print(y)
# there are more functions

func_list = ['a', 'b', ...] # all function's names are stored

chosen_function = random.choice(func_list)
if chosen_function == 'a':
    a(1)
elif chosen_function == 'b':
    b(2)
# elif continues...

I want to eliminates all if statements so that whenever I add new functions in func_list, I do not need to modify the if statements.

However, I can not contain the function itself in the list as the actual functions have randomness in them so I need to call it after the function is sampled.

"However, I can not contain the function itself in the list as the actual functions have randomness in them so I need to call it after the function is sampled." Could you clarify that? — ssp
– ssp, Commented Jan 7, 2021 at 5:42
You won't call a function by merely including it in a list. func_list = [a, b] would work exactly the same way as your if block. — Selcuk
– Selcuk, Commented Jan 7, 2021 at 5:43

Chayim Friedman · Accepted Answer · 2021-01-07 05:47:39Z

3

The answer of @Jarvis is the right way to go, but for completeness sake, I'll show a different way: inspecting the global variables.

def a(x):
    print(x)

def b(y):
    print(y)
# there are more functions

func_list = ['a', 'b', ...] # all function's names are stored

chosen_function = random.choice(func_list)

globals()[chosen_function](x)

answered Jan 7, 2021 at 5:47

Chayim Friedman

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Comments

Jarvis · Accepted Answer · 2021-01-07 05:49:32Z

3

Why not use a dictionary?

def a(x):
    print(x)

def b(y):
    print(y)

func_dict = {'a': a, 'b': b}

Call it like this:

x = 3 # your int input
func_dict['a'](x) # calls a(x)
func_dict['b'](x) # calls b(x)

If you want to directly specify arguments, you can do so in the dictionary like this:

func_dict = {
    'a': lambda: a(1),
    'b': lambda: b(2)
}

Call the default methods like:

func_dict['a']() # calls a(1)
func_dict['b']() # calls b(2)

edited Jan 7, 2021 at 5:49

answered Jan 7, 2021 at 5:42

Jarvis

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Comments

Swadhikar · Accepted Answer · 2021-01-07 05:54:21Z

0

You may consider using eval function. In a nutshell, it evaluates a string into a Python snippet.

For example, in the below code, the eval function evaluates any as a string and interprets to python's any built-in function.

>>> eval('any')
>>> <built-in function any>

On similar grounds, you could evaluate the intended function from a string as below.

def function_a(): 
    print('function A')

def function_b(): 
    print('function B')

function_to_run = eval('function_b')    # function_to_run is a callable now
function_to_run()

Result

function B

answered Jan 7, 2021 at 5:54

Swadhikar

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2 Comments

Jarvis Over a year ago

Use of eval should be highly discouraged. It's dangerous and a bad practice.

Swadhikar Over a year ago

@Jarvis agree. How do you think that code in the question is vulnerable to eval?

sammy · Accepted Answer · 2021-01-07 05:41:32Z

-1

you can set the default value in the function itself

def a(x=1):
    print(x)

def b(y=2):
    print(y)


chosen_function = random.choice(func_list)

chosen_function()

answered Jan 7, 2021 at 5:41

sammy

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1 Comment

OneCricketeer Over a year ago

chosen_function is a string, not a callable

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